LeetCode 2 -- Add Two Numbers

You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

在码的时候需要注意各个边界问题，考虑到零输入，单输入，单进位的可能性，反复修改，例如：[0][1]，[5][5]，[1,8][0]

在编程之前要先想好算法以及过程，先谋而后动。

可以用三指针同时进行，代码会简洁很多，不要破坏原有的链表，这是一个很好的习惯。

 1 public class Solution {
 2     public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
 3        int carry =0;
 4  
 5         ListNode newHead = new ListNode(0);
 6         ListNode p1 = l1, p2 = l2, p3=newHead;
 7  
 8         while(p1 != null || p2 != null){
 9             if(p1 != null){
10                 carry += p1.val;
11                 p1 = p1.next;
12             }
13  
14             if(p2 != null){
15                 carry += p2.val;
16                 p2 = p2.next;
17             }
18  
19             p3.next = new ListNode(carry%10);
20             p3 = p3.next;
21             carry /= 10;
22         }
23  
24         if(carry==1) 
25             p3.next=new ListNode(1);
26  
27         return newHead.next;
28     }
29 }

下面的代码的思路是将为空的链表的值设为0：

 1 public class Solution {
 2     public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
 3         
 4         ListNode result = new ListNode(0);
 5         ListNode start = result;
 6         int temp = 0;
 7         ListNode t1 = l1;
 8         ListNode t2 = l2;
 9         int t1v;
10         int t2v;
11         for (; t1 != null || t2 != null;) {
12             result.next = new ListNode(0);
13             result = result.next;
14             if (t1 == null)
15                 t1v = 0;
16             else {
17                 t1v = t1.val;
18                 t1 = t1.next;
19             }
20             if (t2 == null)
21                 t2v = 0;
22             else {
23                 t2v = t2.val;
24                 t2 = t2.next;
25             }
26             result.val = (t1v + t2v + temp) % 10;
27             temp = (t1v + t2v + temp) / 10;
28 
29         }
30         if (temp != 0) {
31             result.next = new ListNode(0);
32             result.next.val = temp;
33         }
34         return start.next;
35 
36     }
37 }